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Fynjy0 [20]
3 years ago
9

For Valentine’s Day, Sally received a helium-filled balloon at a party. On returning home she accidentally left the balloon in t

he car. Later she went to get the balloon and found it was partially deflated. After being in the house for an hour she noticed it was fully inflated again. Explain why this happened.
Physics
1 answer:
Softa [21]3 years ago
8 0

Answer:

If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.

As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.

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Please help do not guess
Georgia [21]

Answer:

358.9 (+/- 0.4) million years ago

Holocene Epoch, of the Quaternary Period

Devonian period

66 million years ago (prox)

521 million years ago

110,000 years ago

NW

North america became more cold when it moved NW

Explanation:

8 0
3 years ago
In the diagram, the liquid is vaporizing at which point?
sasho [114]
E is the vapourising state
3 0
3 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
A scientist measures the growth of a bamboo plant over time. The table below shows the results.
mihalych1998 [28]
A. 1.35 is the number in between 1.2 and 1.5.
3 0
2 years ago
Read 2 more answers
there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f
Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

4 0
3 years ago
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